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Suppose we are looking at a parabola as a conic in $PG(2,\mathbb{K})$ with the line at infinity denoted $\ell_\infty$. I am still working on a previous problem I posted here.

Suppose we have two points $A(a,a^2)$ and $B(b,b^2)$ on the parabola. The slope of the line $AB$ is $b+a$. How does this line intersect the line at infinity? The idea of the exercise is to get the intersection, $L_{AB}$ of $AB$ and $\ell_\infty$, and then take a line from $L_{AB}$ to the origin and see where it intersects the conic. In this case I think that I always get the origin as the result, but I feel like this should not be the case. I know that the line at infinity is tangent to the parabola, does this mean that the new line will be parallel to $AB$?

mandella
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    By definition any line passing through $L_{AB}$ is parallel to $AB$. So the line through $L_{AB}$ and the origin is just the line through the origin parallel to $AB$. In general, this line should meet the parabola at a second point (unless $a=-b$, or one of $A$, $B$ lies at infinity). – jlammy Jan 03 '21 at 14:07
  • So by definition, $AB$ intersects $\ell_\infty$ at a point, then any line through this point would be parallel to $AB$? Even if I have hyperbola, for example? – mandella Jan 03 '21 at 14:31
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    Yes -- what is your definition of the line at infinity/a point at infinity? – jlammy Jan 03 '21 at 14:32
  • I've been thinking of another construction for a group on the parabola, so I am wondering if the two definitions (like the one in my previous post) are equivalent (https://www.researchgate.net/publication/305987450_Groups_associated_with_conics). – mandella Jan 03 '21 at 14:33
  • It doesn't look like the paper actually defines the line at infinity. The standard definition is that we partition the set of all lines into classes, where two lines are in the same class iff they are parallel. Then for each class, we say that the lines in them "meet" at a point at infinity. The line at infinity just comprises of all the points at infinity. – jlammy Jan 03 '21 at 14:40
  • I did not mention the line at infinity either, but I think we could consider $\ell$ as the line at infinity and get similar results. – mandella Jan 03 '21 at 14:44
  • Are you allowed to use homogeneous projective coordinates? – Somos Jan 03 '21 at 14:53
  • Probably yes, although the hint was to look at $A(a,a^2)$ and $B(b,b^2)$ but I am not sure how the calculations then look like. – mandella Jan 03 '21 at 15:00
  • What is the "origin" in your projective plane? – xxxxxxxxx Jan 05 '21 at 20:16
  • Here I have $(0,0)$. – mandella Jan 05 '21 at 20:18

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If you are allowed to use homogeneous projective coordinates, then some calculations are enough to identify the objects involved. Suppose we have two field elements $\,a,b,\,$ and let $\,c:=a+b.\,$ The point coordinates for the points $\,A,B,L_{AB}\,$ are: $$ A = (a, a\,a, 1), \qquad B = (b, b\,b, 1) \qquad \text{ and }\qquad L_{AB} = (1, c, 0). \tag{1}$$ The line coordinates for the lines $\,AB, \ell_{\infty}, \ell_2\,$ ($\ell_2$ is line through origin and $L_{AB}$) are: $$ AB = (-c, 1, a\,b), \qquad \ell_{\infty} = (0, 0, 1) \qquad \text{ and }\qquad \ell_2 = (c, -1, 0). \tag{2}$$ Define the point $\,C := (c, c\,c, 1)\,$ on the parabola and on the line $\,\ell_2.\,$ Check that all of the other incidence relations hold. For example, line $\,AB\,$ contains point $\,A\,$ since $$ -c(a) + 1(a\,a) + a\,b(1) = -(a+b)a + a\,a + a\,b = 0. \tag{3}$$

Projectively, this is equivalent to a circle with a point $\,O\,$ on it regarded as the origin and then two points $\,A\,$ and $\,B\,$ on the circle determine a third $\,C\,$ so that the line $\,AB\,$ is parallel to $\,OC.\,$ This process defines a group operation on the circle points.

Somos
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